# Application of Picard-Lindelöf to determine uniqueness of a solution to an IVP

I am still struggling quite a lot with the Picard-Lindelöf-Theorem (also known as the Cauchy-Lipschitz-Theorem).

Problem: Consider the following IVP with $\alpha \neq 1$ $$\begin{cases} y'&= t|y|^\alpha \\ y(0)&=1 \end{cases}$$ And show that there exists a unique solution $f_\alpha$ on an Interval $I_\alpha$ with $0 \in I_\alpha$, also show that $f_\alpha (t) > 0$ for small $t$

So to use Picard-Lindelöf I want to show that $f(t,y)=t|y|^\alpha$ is Lipschitz-continuous with respect to its second variable. I could do this by computing the partial derivative with respect to $y$ and find an interval on which this expression is bound, then on said hypothetical interval the solution would exist and be unique.

My approach: $$\frac{\partial f}{\partial y}= t\alpha|y|^{\alpha-1} \cdot \frac{y}{\sqrt{y^2}}$$ At which point I already run into trouble. I have made use of the fact that $|x|'= x/\sqrt{x^2}$ because it gives correct results in regard of the direction (when approaching from the left and the right). But the above expression is not defined at $y=0$.

So I thought if I could come up with an argument that shows why $y \neq 0$, then at least this problem would be out of the way.

Assume that $y=0$ then $y'=0=f(t,y) \implies \frac{\partial f}{\partial y}=0$ which is bound on $\mathbb{R} \implies f$ is Lipschitz continuous with respect to $y$ but we have $y(0)=1$ and thus $$y'=0 \implies y=c$$ So $y$ is a constant function and $y(0)=1$ means that $c=1 \implies y=1 \neq 0$ which shows that $y$ can not be the 'zero' function (if that exists).

Questions:

• Is the above argumentation correct?
• How can I find the desired interval $I_\alpha$ with $0 \in I_\alpha$?

Update: I tried to continue a bit on my own and ignore the Picard-Lindelöf part and just focus on solving the IVP, apparently the step I have taken above was necessary so I can divide by $|y|^\alpha$, however I will land at $$\frac{dy}{|y|^\alpha}=tdx$$ So it seems like I have to come up with a condition that guarantees me that $y$ is positive in some Intervall.

It says: consider the problem $y'=f(t,y)$ with $y(t_0)=y_0$, and suppose $(t_0,y_0)$ has a neighborhood (a rectangle $R$ in $ty$-plane) in which $f$ is Lipschitz in $y$ and continuous in $t$. Then there is an interval $(t_0-\delta,t_0+\delta)$ in which a solution exists, is unique, and stays within the aforementioned rectangle.
In your case, $(t_0,y_0)=(0,1)$. Take $R$ to be $[-1,1]\times [1/2,3/2]$; all that matters is that we stay away from $y=0$. Check that $f$ satisfies the above properties on $R$; this isn't hard especially because $|y|=y$ (we have positive values of $y$ only). Apply the theorem. Done.
Nothing in the problem statement says that you have to produce an explicit interval of existence/uniqueness. But if you want to do this, it's best to use the optimized version which gives the interval of existence/uniqueness as $|t-t_0|\le \min(a,b/M)$. Here $a$ and $b$ describe the rectangle $R$ that is ours to choose; with my choice $a=1$ and $b=1/2$. The number $M$ is the maximum of $|f|$ on said rectangle, which turns out to be $(3/2)^\alpha$. So, in the interval $|t|\le 2^{-1}(2/3)^{\alpha}$ the solution exists and is unique.